- #1
titaniumx3
- 53
- 0
Hi,
I have this question which asks for a 2*pi periodic function on the reals, that is integrable on [-pi, pi] but fails to satisfy the mean value theorem for integrals.
The question also says to help answer the above question, you may wish to show that the function:
g: [0,1] -> reals given by,
g(x) = sin(1/x) if x is non-zero
g(x) = 0 if x=0Now I've shown that the above function is true, but I have no idea how to show that it fails to satisfy the mean value theorem or how it relates to a function that is 2*pi periodic and integrable on the given interval.
Please help!
I have this question which asks for a 2*pi periodic function on the reals, that is integrable on [-pi, pi] but fails to satisfy the mean value theorem for integrals.
The question also says to help answer the above question, you may wish to show that the function:
g: [0,1] -> reals given by,
g(x) = sin(1/x) if x is non-zero
g(x) = 0 if x=0Now I've shown that the above function is true, but I have no idea how to show that it fails to satisfy the mean value theorem or how it relates to a function that is 2*pi periodic and integrable on the given interval.
Please help!